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# By abc we denote a three digit number with digits a, b, and c. Is abc

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Math Expert
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By abc we denote a three digit number with digits a, b, and c. Is abc  [#permalink]

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18 Mar 2015, 05:19
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55% (hard)

Question Stats:

61% (01:04) correct 39% (00:47) wrong based on 199 sessions

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By abc we denote a three digit number with digits a, b, and c. Is abc divisible by 3?

(1) The product of (a) times (b) is a number divisible by 3
(2) c = 3

Kudos for a correct solution.

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Joined: 08 Feb 2015
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By abc we denote a three digit number with digits a, b, and c. Is abc  [#permalink]

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18 Mar 2015, 10:04
1
Bunuel wrote:
By abc we denote a three digit number with digits a, b, and c. Is abc divisible by 3?

(1) The product of (a) times (b) is a number divisible by 3
(2) c = 3

Kudos for a correct solution.

for this to be correct, a+b+c must be divisble by 3.

1: this means one/two of a and b is divisible by 3. it tells nothing about other two/one. INSUFFICIENT.

2: tells nothing about a and b. INSUFFICIENT.

combine both:

abc can be divisble if abc = 333
abc CAN NOT be divisible if abc 233

Ans: E
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Re: By abc we denote a three digit number with digits a, b, and c. Is abc  [#permalink]

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18 Mar 2015, 11:34
1
(1) a*b is divisible by 3
Let a = 3, b = 1 then if abc = 312, abc is divisible by 3 if, abc = 313 then it is not divisible by 3. INSUFFICIENT
(2) c = 3
This is clearly INSUFFICIENT. Since, a & b are unknown.

(1) & (2) together also are INSUFFICIENT as seen. Hence, Answer is E.
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Re: By abc we denote a three digit number with digits a, b, and c. Is abc  [#permalink]

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18 Mar 2015, 22:02
1
ABC is divisible by 3 => A+B+C needs to be divisible by 3

(1) The product of (a) times (b) is a number divisible by 3 => At least a or b is divisible by 3 => INSUFFICIENT
(2) c = 3 => INSUFFICIENT

(1) + (2) => At least 2 of a,b,c are divisible by 3, BUT NOT SURE a+b+c is divisible by 3

Math Expert
Joined: 02 Sep 2009
Posts: 49303
Re: By abc we denote a three digit number with digits a, b, and c. Is abc  [#permalink]

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23 Mar 2015, 06:15
Bunuel wrote:
By abc we denote a three digit number with digits a, b, and c. Is abc divisible by 3?

(1) The product of (a) times (b) is a number divisible by 3
(2) c = 3

Kudos for a correct solution.

MAGOOSH OFFICIAL SOLUTION:

Remember, the divisibility rule: if the sum of all the digits in a large number is divisible by 3, then the number itself is divisible by three.

The problem with statement #1, for any given values of a & b that work, the one’s digit, c, could be anything. For example, a = 1, and b = 3 satisfy this condition, so that would be all the numbers in the 130’s: {130, 131, 13, 133, 134, 135, 136, 137, 138, 139}. Clearly, some of these are divisible by 3, namely {132, 135, 138} and the others aren’t. This statement allows for multiple answers, so this statement, alone and by itself, is not sufficient.

The problem with statement #2 is that we now know the one’s digit, but the other two digits could be anything. Think about the first few possible numbers of this form: {103, 113, 123, 133, 143, 153, 163, 173, 183, 193, 203}. Of those, only (123, 153, 183) are divisible by three, and the others aren’t. In fact, you would be expected to know this, but as it happens, {103, 163, 173, 193} are prime numbers, not divisible by anything! This statement also allows for multiple answers, so this statement, alone and by itself, is not sufficient.

Combined statements: now, we have to obey both constraints. One number that obeys both constraints is 133, which is not divisible by 3. Another number that obeys both constraints is 333, which is divisible by 3. Even with both statements in place, we can find numbers that produce either a “yes” or a “no” to the prompt. Even together, the statements are not sufficient.

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Re: By abc we denote a three digit number with digits a, b, and c. Is abc  [#permalink]

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15 Mar 2016, 10:15
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Joined: 20 Jun 2016
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Re: By abc we denote a three digit number with digits a, b, and c. Is abc  [#permalink]

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20 Aug 2017, 22:38
The problem with statement #1, for any given values of a & b that work, the one's digit, c, could be anything. For example, a = 1, and b = 3 satisfy this condition, so that would be all the numbers in the 130's: {130, 131, 13, 133, 134, 135, 136, 137, 138, 139}. Clearly, some of these are divisible by 3, namely {132, 135, 138} and the others aren’t. This statement allows for multiple answers, so this statement, alone and by itself, is not sufficient.

The problem with statement #2 is that we now know the one's digit, but the other two digits could be anything. Think about the first few possible numbers of this form: {103, 113, 123, 133, 143, 153, 163, 173, 183, 193, 203}. Of those, only (123, 153, 183) are divisible by three, and the others aren’t. In fact, you would NOT be expected to know this, but as it happens, {103, 163, 173, 193} are prime numbers, not divisible by anything! This statement also allows for multiple answers, so this statement, alone and by itself, is not sufficient.

Combined statements: now, we have to obey both constraints. One number that obeys both constraints is 133, which is not divisible by 3. Another number that obeys both constraints is 333, which is divisible by 3. Even with both statements in place, we can find numbers that produce either a "yes" or a "no" to the prompt. Even together, the statements are not sufficient.

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Re: By abc we denote a three digit number with digits a, b, and c. Is abc  [#permalink]

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28 Oct 2017, 07:49
For divisibility of 3, a+b+c should be a multiple of 3.

Statement 1 is not sufficient as the number can be 331 or 333.

Statement 2 is also not sufficient as the number can be 223 or 333

Statement 1 +2 also not sufficient as number can be 133 or 333

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Re: By abc we denote a three digit number with digits a, b, and c. Is abc &nbs [#permalink] 28 Oct 2017, 07:49
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# By abc we denote a three digit number with digits a, b, and c. Is abc

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